Answer
The angular speed is $1.04~rad/s$
Work Step by Step
We can find the rotation radius for each car:
$r = 6.00~m+(4.25~m)~sin~45^{\circ}$
$r = 9.0~m$
Let $m$ be the mass of one car. Let $T$ be the tension in the cable attached to one car. The vertical component of tension is equal in magnitude to the car's weight:
$T~cos~45^{\circ} = mg$
$T = \frac{mg}{cos~45^{\circ}}$
The horizontal component of the tension provides the centripetal force to keep the car moving in a circle. We can find the angular speed $\omega$:
$T~sin~45^{\circ} = m~\omega^2~r$
$(\frac{mg}{cos~45^{\circ}})~sin~45^{\circ} = m~\omega^2~r$
$g~tan~45^{\circ} = \omega^2~r$
$\omega^2 = \frac{g~tan~45^{\circ}}{r}$
$\omega = \sqrt{\frac{g~tan~45^{\circ}}{r}}$
$\omega = \sqrt{\frac{(9.80~m/s^2)~tan~45^{\circ}}{9.0~m}}$
$\omega = 1.04~rad/s$
The angular speed is $1.04~rad/s$
